Question: What is the extraneous solution to these equations? $\dfrac{x^2}{x - 4} = \dfrac{x + 20}{x - 4}$
Explanation: Multiply both sides by $x - 4$ $ \dfrac{x^2}{x - 4} (x - 4) = \dfrac{x + 20}{x - 4} (x - 4)$ $ x^2 = x + 20$ Subtract $x + 20$ from both sides: $ x^2 - (x + 20) = x + 20 - (x + 20)$ $ x^2 - x - 20 = 0$ Factor the expression: $ (x + 4)(x - 5) = 0$ Therefore $x = -4$ or $x = 5$ The original expression is defined at $x = -4$ and $x = 5$, so there are no extraneous solutions.